r 1 = P(q 1-20)=(-2x+80)[(x+30)-20]，

=-x2+20x+800( 1≤x≤20, where x is an integer),

R2=P(Q2-20)=(-2x+80)(45-20)，

=-50x+2000 (2 1 ≤ 30, x is an integer);

(2) When 1≤x≤20 and x is an integer,

∫r 1 =-(x- 10)2+900，

When x= 10, the maximum value of R 1 is 900.

When 2 1≤x≤30 and x is an integer,

∵R2=-50x+2000, -50 < 0, R2 decreases with the increase of x,

∴ When x=2 1, the maximum value of R2 is 950.

∵950>900,

∴ When x=2 1, that is, on 2 1 day, the daily sales profit is the largest, and the maximum value is 950 yuan. Comments: This question needs to be read repeatedly, and the function relationship is established according to the basic equivalence relationship in marketing problems, and the piecewise function is listed according to the time period. Then, the maximum values of the two functions are obtained by combining the range of independent variables, and the conclusion is drawn.